Lecture 8: Equivalence Relations

Transcription

1 Lecture 8: Equivalence Relations 1 Equivalence Relations Next interesting relation we will study is equivalence relation. Definition 1.1 (Equivalence Relation). Let A be a set and let be a relation on A. The relation is an equivalence relation if it is reflexive, symmetric, and transitive. Example 1.2. Following are some of the examples of equivalence relations. 1. Let P be set of all people. Then, we can define an equivalence relation = {(x, y) P P : x and y have same age}. 2. We can define an equivalence relation on real numbers as {(x, y) R R : x = y}. 3. Let A be a non-empty set. We can define an equivalence relation on power set of A as {(S, T ) P(A) P(A) : S = T }. Definition 1.3. Let A be a non-empty set and be an equivalence relation on A. 1. The relation classes of A with respect to are called equivalence classes, and denoted [x] for all x A. 2. The quotient set of A and is the set { [x] : x A} of all equivalence classes of A with respect to, and is denoted by A/. Example 1.4. Let P be the set of all people, and let be the relation on P defined by x y if and only if x and y are the same age (in years). If person x is 19 years old, then the equivalence class of x is the set of all 19-year olds. Each element of the quotient set P/ is itself a set, where there is one such set consisting of all 1-year-olds, another consisting of all 2-year olds, and so on. Although there are billions of people in P, there are fewer than 125 elements in P/, because no currently living person has reached the age of

2 Theorem 1.5. Let A be a non-empty set, and let be an equivalence relation on A. i) Let x, y A. If x y, then [x] = [y]. If x y, then [x] [y] =. ii) [x] = A. x A Proof. Let A be a non-empty set, and let be an equivalence relation on A. i) Let z [x]. Then z x by assumption and x y by hypothesis. By transitivity of, we have z y, and hence the z [y]. Therefore, we conclude [x] [y]. Similarly, we can show [y] [x]. We prove the second part by contradiction. Let x y, and z [x] [y]. Then, z x and z y. By transitivity and symmetry of, we get x y. Therefore, we have a contradiction. ii) By definition, [x] A for all x A. Hence, x A [x] A. Now, let q A. Then, q q by reflexivity. Therefore, q [q] [x]. Hence, A [x]. x A x A Corollary 1.6. Let A be a non-empty set, let be an equivalence relation on A. Let x, y A. Then [x] = [y] iff x y. Since equivalence relations are disjoint for unrelated elements, quotient sets separates a set into disjoint union of equivalence classes. Definition 1.7 (Partition). Let A be a non-empty set. A partition of A is a family D of non-empty subsets of A such that 1. if P, Q D and P Q, then P Q =, and 2. P = A. P D Example 1.8. We look at some examples of partitions. 1. Let D = {E, O}, where E and O are set of even and odd integers respectively. Then, D is a partition of integers Z. 2. Collection of sets C = {[n, n + 1) : n Z} is a partition of R. 3. Collection of sets G = {(n 1, n + 1)} is not a partition of R because it is not pairwise disjoint. We have two sets (-1,1) and (0, 2) in C, that are not disjoint. In fact, ( 1, 1) (0, 2) = (0, 1). 2

3 From Theorem 1.5, and definition of partitions, we have the following corollary. Corollary 1.9. Let A be a non-empty set, and let be an equivalence relation on A. Then A/ is a partition of A. Definition Let A be a non-empty set. Let E(A) denote the set of all equivalence relations on A. Let T A denote the set of all partitions of A. Example Let A = {1, 2, 3}. Then T A = {D 1, D 2, D 3, D 4, D 5 }, where D 1 = {{1}, {2}, {3}}, D 2 = {{1, 2}, {3}}, D 3 = {{1, 3}, {2}}, D 4 = {{2, 3}, {1}}, D 5 = {{1, 2, 3}}. Further, we see that E(A) = {R 1, R 2, R 3, R 4, R 5 }, where these equivalence relations are defined by the sets R 1 = {(1, 1), (2, 2), (3, 3)}, R 2 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}, R 3 = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}, R 4 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}, R 5 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}. It is easy to see that each of the relations {R i, i = 1,..., 5} listed above is an equivalence relation on A. It is interesting to note that number of equivalence relations on a set, and number of partitions are same for the above example. It turns out that this is not mere coincidence. Definition Let A be a non-empty set. Let Φ : E(A) T A be defined as follows. If is an equivalence relation on A, let Φ( ) be the family of sets A/. Let Ψ : T A E(A) be defined as follows. If D is a partition of A, let Ψ(D) be the relation on A defined by x Ψ(D) y iff there is some P D such that x, y P for all x, y A. Lemma Let A be a non-empty set. The functions Φ and Ψ in the above definition are well defined. Proof. To prove the lemma, we need to show the following two things. 3

4 1. For any equivalence relation on A, the family of sets Φ( ) is a partition of A. 2. For any partition D of A, the relation Ψ(D) is an equivalence relation on A. First part follows from the definition of Φ and Corollary 1.9. We will show the second part. Any partition D = {P i : i {1,..., n}} for some n. From definition of partition, it implies A = n P i and P i P j = φ for all i j. Therefore, i=1 Ψ(D) = {(x, y) : there is some i {1,..., n} D such that x, y P i }. We will show that Ψ(D) is an equivalence relation on A. Symmetry: Let x Ψ(D) y. Then we can find P D such that x, y P. Hence, y Ψ(D) x. Reflexivity: Let x A. Then x P for some P D. Hence, x Ψ(D) x. Transitivity: Let x Ψ(D) y and y Ψ(D) z. Then, x, y P for some P D and y, z Q for some Q D. Since, y P Q and P and Q are disjoint for P Q, it follows that P = Q. This implies x, y, z P. Hence, x Ψ(D) z. Example We look at some examples of equivalence classes and related functions Φ, and partitions and related function Ψ. 1. Let be the relation on R 2 defined by (x, y) (z, w) iff y x = w z, for all (x, y), (z, w) R 2. It can be verified that is an equivalence relation. We want to describe the partition Φ( ) of R 2. Let (x, y) R 2. Then Let c = y x, then [(x, y)] = {(z, w) R 2 w z = y x}. [(x, y)] = {(z, w) R 2 w = z + c}, which is just a line in R 2 with slope 1 and y-intercept c. Hence, Φ( ) is collection of all lines in R 2 with slope 1. 4

5 2. Let C = {[n, n+1) : n Z} be a partition of R. Then, we have corresponding family of equivalence relation classes, defined as Ψ(C) = {(x, y) R R : x, y [n, n + 1) for some n Z} = {(x, y) R R : x = y }. 3. For partitions D i and equivalence relations R i defined in Example 1.11, we have Φ(R i ) = D i and Ψ(D i ) = R i for all i {1,..., 5}. Theorem Let A be a non-empty set. Then the functions Φ and Ψ are inverses of each other, and hence both are bijective. Proof. We need to show that Ψ Φ = 1 E(A) and Φ Ψ = 1 TA. We will do this in following steps. 1. First, we prove that Ψ Φ = 1 E(A). Let E(A) be an equivalence relation on A. Let = Ψ(Φ( )). We will show that =, and it will then follow that Ψ Φ = 1 E(A). We show two relations are equal by set equality. Let D = Φ( ), so that = Ψ(D). (a) First, we show that. Let x, y A such that x y. Then, by the definition of Ψ there is some P D such that x, y P. By the definition of Ψ, we know that P is an equivalence class of, so that P = [q] for some q A. Then q x and q y, and by the symmetry and transitivity of it follows that x y. Hence,. (b) Now, we show that. Let x y in A. Then, y [x]. By the reflexivity of, we know that x [x]. From definition of Φ we have [x] D. Hence, by the definition of Ψ, it follows that x y. Hence,. 2. Second, we prove that Φ Ψ = 1 TA. Let D T A be a partition of A. Let F = Φ(Ψ(D)). We will show that F = D, and it will then follow that Φ Ψ = 1 TA. Let = Ψ(D), so that F = Ψ( ). (a) Let B F. Then by the definition of Φ we know that B is an equivalence class of, so that B = [z] for some z A. Because D is a partition of A, then there is a unique P D such that z P. Let w A. Then by the definition of Ψ we see that z w if and only if w P. It follows that w [z] if and only if w P, and hence P = [z]. Hence B = [z] = P D. Therefore F D. (b) Let C D. Let y C. As before, it follows from the definition of Ψ that C = [y]. Therefore, by the definition of Φ we see that C Φ( ) = F. Hence, D F. 5

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